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\title{Introduction to Quantum Mechanics\\Lecture 7}
\author{Brian Greene\footnote{These lecture notes were TeX'd by Nilay Kumar}} % ADD YOUR NAME HERE IF YOU WORK ON THIS LECTURE's NOTES
\date{October 4, 2011}

\maketitle

So far, we have been analyzing the results of certain quantum mechanical concepts that are relevant in the microscopic world. At this point,
one might ask how classical physics, namely, Newton's second law, arises from the quantum formalism that we have set up so far. The goal of
this lecture is to introduce the expectation value and its meaning in QM, and to see how classical mechanics, in a sense, simply an
approximate regime of quantum mechanics.

\section{Expectation values}
Suppose we have an ensemble of systems, each in the state $\psi_0$. In other words, imagine that we have $N$ independent, non-interacting
boxes, each of which identically contain a particle in the state $\psi_0$. Suppose, at any given moment in time, we perform a measurement on
each of the systems. According to the probabilistic interpretation of quantum mechanics, we will not necessarily find that each particle has
collapsed to and identical state $\psi_f$ - in general, each of the states will be different, and thus will yield different values for the
observable under consideration. For example, we might measure the particle in the first box to be located at $x=17$, whereas we measure the
particle in the second box to be at $x=\pi$. As a whole, however, the results obtained from simultaneous measurements performed on this
ensemble should form some distribution that roughly matches $\psi$ (with the approximation becoming exact as $n\to\infty$).

This motivates us to consider what the average of the distribution of the results might be. This average is called the expectation value and
is denoted, for example, as $\langle x\rangle$ for the expectation value of position. It is defined, quite intuitively, as 
\[\langle q \rangle=\sum_i^N q_i\cdot\text{prob}(q_i),\]
the average of the results weighted by the probability of obtaining each result. To apply this to the position of a particle, we first note
that $x$ is a continuous variable and that the probability of finding the particle at some location $x$ (and fixed time $t_0$) is given by
$|\psi(x)|^2\;dx$. The
above sum now becomes an integral\footnote{Unless expressly stated otherwise, all integrals are evaluated from $-\infty$ to $\infty$.}:
\begin{equation}
    \langle x\rangle =\int dx\;\psi^*(x,t_0)x\psi(x,t_0)
    \label{eq:expectx}
\end{equation}
Note that we have sandwiched the $x$ in the norm of the wavefunction - we will see later why this convention is useful.

It is important to realize that the expectation value \textit{does not} represent the most likely possibility - in fact, it can take on a
value $x$ where $\psi(x)$ is in fact 0. A simple example is the case of an even function $\psi_e$ that is zero at $x=0$. Intuitively, it is
just as probable that the particle will be found to the left of the origin as it will to the right. Consequently, the average value of
$\langle x\rangle$ is just $x=0$.

\section{Expectation values of momentum}
We now have a method of calculating the expectation value of the position of a particle with wavefunction $\psi$. Can we do the same with
its momentum? Well, based on our intuition about the relationship between $\psi$ and $\phi$, one would hope that
\begin{equation}
    \langle k\rangle=\int dk\;\phi^*(x,t_0)k\phi(x,t_0)
    \label{eq:expectk}
\end{equation}
How can we justify this? Certainly, any reasonable person would expect that
\begin{equation*}
    \begin{split}
        \langle p\rangle&=m\frac{d\langle x\rangle}{dt}\\
        &=m\frac{d}{dt}\int dx\;\psi^*x\psi\\
        &=m\int dx\;x\left( \frac{d\psi^*}{dt}\psi+\frac{d\psi}{dt}\psi^*\right)
    \end{split}
\end{equation*}
Using the Schr\"{o}dinger equation to find the time-derivatives of $\psi$ and $\psi^*$, and by integrating by parts twice, we find that
\begin{equation}
    \langle p\rangle=\int dx\;\psi^*\left( -i\hbar\frac{\partial}{\partial x} \right)\psi
    \label{eq:expectp}
\end{equation}
Note that the term sandwiched between $\psi^*$ and $\psi$ is no longer a variable, but is now a (linear) operator. We will revisit the
significance of this distinction in later lectures. Now let us check this result against what we know about the Fourier relationship
between $\psi$ and $\phi$:
\begin{equation*}
    \begin{split}
        -i\hbar\frac{\partial\psi}{\partial x}&=-i\hbar\frac{\partial}{\partial x}\frac{1}{\sqrt{2\pi}}\int dk\;\phi(k,t)e^{ikx}\\
        &=\frac{\hbar k}{\sqrt{2\pi}}\int dk\;\phi(x,t)e^{ikx}
    \end{split}
\end{equation*}
Using this relationship together with Eq.~(\ref{eq:expectp}), we can write,
\begin{equation*}
    \begin{split}
        \langle p\rangle&=\langle \hbar k\rangle=\frac{1}{2\pi}\int dk\;dk'\;dx\;\phi^*(k',t)(\hbar k)e^{i(k-k')x}\phi(k,t)
    \end{split}
\end{equation*}
Recognizing the $x$ integral as a delta function, we can collapse the $k'$ prime integral, yielding 
\begin{equation*}
    \langle p\rangle=\int dk\;\phi^*(k,t)(\hbar k)\phi(k,t),
\end{equation*}
which agrees with our intuitive guess from before - Eq.~(\ref{eq:expectk}).

\section{Recovering Newton's law}
Now that we have an expression for the expectation values of $p$, we can extract Newton's law from the formalism of quantum mechanics.
First, note that Newton's law can be written as
\begin{equation}
    F=\frac{dp}{dt}=-\frac{\partial V}{\partial x}.
    \label{eq:newton}
\end{equation}
This form motivates us to take the time derivative of $\langle p \rangle$:
\begin{equation*}
    \begin{split}
        \frac{d\langle p\rangle}{dt}&=\frac{d}{dt}\int dx\;\psi^*\left[ -i\hbar\frac{\partial}{\partial x} \right]\psi\\
        &=\int dx \frac{\partial\psi^*}{\partial t}\left[-i\hbar\frac{\partial}{\partial x}\right]\psi+\psi^*\left[
        -i\hbar\frac{\partial}{\partial x} \right]\frac{\partial\psi}{\partial t}
    \end{split}
\end{equation*}
Using Schr\"{o}dinger's equation to determine the time-derivatives of the wavefunction, the above equation becomes
\begin{equation*}
    \frac{d\langle p\rangle}{dt}=\int dx\;\frac{-\hbar^2}{2m}\frac{\partial^2\psi^*}{\partial x^2}\frac{\partial\psi}{\partial
    x}+V\psi^*\frac{\partial\psi}{\partial
    x}+\frac{\hbar^2}{2m}\psi^*\frac{\partial^3\psi}{\partial x^3}-\psi^*\frac{\partial}{\partial x}(V\psi)        
\end{equation*}
Integrating the first term twice yields the negative of the third term, and thus both vanish. We are then left with
\begin{equation*}
    \frac{d\langle p\rangle}{dt}=\int dx\;V\psi^*\frac{\partial\psi}{\partial
    x}-\psi^*V\frac{\partial \psi}{\partial x}-\psi^*\frac{\partial V}{\partial x}\psi,
\end{equation*}
in which the first two terms cancel. Finally, we are left with
\begin{equation*}
    \frac{d\langle p\rangle}{dt}=-\int dx\;\psi^*\frac{\partial V}{\partial x}\psi=\langle -\frac{\partial V}{\partial x}\rangle
\end{equation*}
by our definition of expectation values - i.e. the average value of the force over the $x$-axis. This is Ehrenfest's theorem - that
expectation values follow Newton's law. However, this is not fully Newton's law yet - somehow, we need to get expectation values to equal
the \textit{actual} values. We can speak of an actual value in this instance, because in the classical regime, wavefunctions are sharply peaked
about such a value - the expectation value when using a sharply peaked wavefunction will simply pick out the actual value. But what's to
keep the wavefunction peaked over time? Can't it spread out? No. Due to the large masses that we typically deal with in the classical
regime, wavefunctions spread out extremely slowly, and so for any given time in the evolution of our system, we can write, approximately,
\[\frac{dp}{dt}=-\frac{\partial V}{\partial x}.\]
We now see how classical mechanics is just an approximation of quantum mechanics, which is incredibly useful (and accurate) when
wavefunctions are sharply peaked through time.


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